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Z and T' parameters with dependent sources

Short answer

Z-parameters: Z = [12 5; −6 −4] Ω. Inverse transmission parameters: T' = [−4/5 −18/5; 1/5 12/5]. The network is not symmetrical (z11 ≠ z22) and not reciprocal.

Find the Z-parameter and hence T'-parameter for the network shown in figure below also check if network is symmetrical. [Figure: port 1 (V1, I1) followed by a 10 Ω series resistor, then a dependent voltage source 3I2 in series, to a middle node X. A 2 Ω resistor with voltage V3 across it (+ at top) is connected from X to the bottom wire, and a 2 Ω series resistor connects X to port 2 (V2, I2). A dependent current source 2V3 (arrow pointing toward port 1) is connected from the node after the 10 Ω resistor to the port 2 terminal, across the dependent voltage source and the 2 Ω series resistor.]

Z-parameters: V1 = z11 I1 + z12 I2, V2 = z21 I1 + z22 I2. T'-parameters (inverse transmission): V2 = A' V1 − B' I1, I2 = C' V1 − D' I1.

Reading the figure (assumptions). From port 1: a 10 Ω series resistor to node N, then the current-controlled voltage source 3I2 in series (+ on the port 1 side), reaching node X. A 2 Ω shunt resistor from X to the bottom wire carries V3 = VX (+ at the top). A 2 Ω series resistor goes from X to port 2. The current source 2V3, with its arrow pointing towards port 1, connects the port 2 terminal to node N (just before the 3I2 source).

Equations (all currents in A, resistances in Ω)

(1) V1 = VN + 10 I1                       (10 Ω drop)
(2) VX = VN − 3 I2                        (3I2 source, + towards port 1)
(3) KCL at N: I1 + 2VX = Isrc             (2V3 flows into N; Isrc goes through the 3I2 source)
(4) KCL at X: Isrc = VX/2 + (VX − V2)/2 = VX − V2/2
(5) KCL at port 2 node: I2 + (VX − V2)/2 = 2VX

Solving

From (3) and (4): I1 + 2VX = VX − V2/2, so VX = −I1 − V2/2. Equation (5) gives I2 = 1.5 VX + V2/2. Substituting VX = −I1 − V2/2: I2 = −1.5 I1 − 0.75 V2 + 0.5 V2 = −1.5 I1 − 0.25 V2, so V2 = −6 I1 − 4 I2. Then VX = −I1 − V2/2 = 2 I1 + 2 I2. From (1) and (2): V1 = VX + 3 I2 + 10 I1 = 12 I1 + 5 I2.

z11 = 12 Ω    z12 = 5 Ω
z21 = −6 Ω    z22 = −4 Ω
Δz  = z11 z22 − z12 z21 = −48 + 30 = −18

T'-parameters

Eliminate I2 from the Z equations: I2 = (V1 − z11 I1)/z12.

A' = z22/z12 = −4/5          B' = Δz/z12 = −18/5
C' = 1/z12   = 1/5           D' = z11/z12 = 12/5

Symmetry check

A network is symmetrical if z11 = z22 (and it is reciprocal). Here z11 = 12 Ω ≠ z22 = −4 Ω, and also z12 = 5 ≠ z21 = −6 (the dependent sources make it non-reciprocal).

Answer: Z = [12 5; −6 −4] Ω; T' = [A' B'; C' D'] = [−4/5 −18/5; 1/5 12/5]; the network is not symmetrical (and not reciprocal).

✓ Answer checked by Claude Opus 5.5 (high effort)

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